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antoniya [11.8K]
2 years ago
7

Consider the system consisting of two blocks connected by a rope and a pulley. The coefficient of static friction between the ra

mp and the block on the ramp is 0.36. You may assume the tension in the string connecting the two blocks is the same for all parts of the string. The two arrows define a coordinate system for the block on the ramp. The x axis is parallel to the ramp, the y axis is perpendicular to the ramp. The angle the ramp makes with the horizontal ground is 35 degrees. The block on the ramp has a weight of 13 Newtons.
Required:
What is the heaviest weight that can be hung below the pulley without the block on the ramp moving?
Physics
1 answer:
RSB [31]2 years ago
5 0

Answer:

1.2 kg

Explanation:

Let UP ramp be the positive direction

                                                  F = ma

   T     -   Wt ||      -         Ff            = m(0)

  mg   -  Μgsinθ -      μΜgcosθ    = 0

m(9.8) - 13sin35 - 0.36(13)cos35 = 0        

                                                 m = 13(sin35 + 0.36cos35) / 9.8

                                                 m = 1.15205... ≈ 1.2 kg

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if quasars often resemble little blue stars, what was it about them that so surprised astronomers when they were dis­covered?
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1 year ago
A 1400.0 kg car crests a 3200.0 m pass in the mountains and briefly comes to rest. The car descends 1000 m before climbing and c
rjkz [21]

Answer:

a) v = 88.54 m/s

b) vf = 26.4 m/s

Explanation:

Given that;

m = 1400.0 kg

a)

by using the energy conservation

loss in potential energy is equal to gain in kinetic energy

mg × ( 3200-2800) = 1/2 ×m×v²

so

1400 × 9.8 × 400 = 0.5 × 1400 × v²

5488000 = 700v²

v² = 5488000 / 700

v² = 7840

v = √7840

v = 88.54 m/s

b)

Work done by all forces is equal to change in KE

W_gravity + W_non - conservative = 1/2×m×(vf² - vi²)

we substitute

1400 × 9.8 × ( 3200-2800) - (5 × 10⁶) = 1/2 × 1400 × (vf²  -0 )

488000 = 700 vf²

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3) Calculate the kinetic energy of a 7 kg mass traveling at a velocity of 4 m/sec.
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▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

Let's solve ~

Given terms :

  • Mass (m) = 7 kg

  • velocity (v)= 4 m/s

The formula to find kinetic Energy is ~

\boxed{ \boxed{ \sf{ \frac{1}{2}  m{v}^{2} }}}

Now, apply the formula according to given situation

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: \dfrac{1}{2}  \times 7 \times ( {4)}^{2}

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \: \dfrac{1}{2}  \times 7 \times 16

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:7 \times 8

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:56 \:  \: joules

Therefore, the kinetic Energy of the car is 56 joules

4 0
2 years ago
Read 2 more answers
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